Question

A charge of −31 μC is distributed uniformly throughout a spherical volume of radius 8 cm. Determine the electric field due to this charge at a distance of (a) 1.7 cm, (b) 5.5 cm, and (c) 24 cm from the center of the sphere

Answer 1

Answer:

(a) $E = -9.3 \times 10^6~N/C$

(b) $E = -3.01 \times 10^7~N/C$

(c) $E = -4.86 \times 10^6~N/C$

Explanation:

Gauss' Law can be applied to find the E-field.

E-field should be calculated separately for the inside and outside.

Inside:

$\int {\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The integral in the left-hand side of the above equation is equal to the electric flux through the imaginary spherical shell that we have put inside the sphere. And Q in the right-hand side is the enclosed charge inside this imaginary surface.

The enclosed charge can be calculated as follows

$\frac{Q_{total}}{V_{total}} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi R^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{R^3}$

where R is the radius of the total sphere, and r is the radius of the imaginary shell.

Applying Gauss' Law gives for any point inside the sphere

$E4\pi r^2 = \frac{Qr^3}{\epsilon_0 R^3}\\E = \frac{Qr}{4\pi\epsilon_0 R^3}$

Outside:

The imaginary surface for the Gauss' Law should be put outside the sphere.

Now, the enclosed charge is equal to the total charge of the sphere, Q.

$\int Eda = \frac{Q_{encl}}{\epsilon_0}\\E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$

Now, the question can be solved.

(a) r = 0.017 m. Inside the sphere.

$E = \frac{Qr}{4\pi\epsilon_0 R^3} = \frac{(-31\times 10^{-6})(0.017)}{4\pi(8.8\times 10^{-12})(0.08)^3} = -9.3 \times 10^6~N/C$

(b) r = 0.055 m. Inside the sphere.

$E = \frac{Qr}{4\pi\epsilon_0 R^3} = \frac{(-31\times 10^{-6})(0.055)}{4\pi(8.8\times 10^{-12})(0.08)^3} = -3.01 \times 10^7~N/C$

(c) r = 0.24 m. Outside the sphere.

$E = \frac{Q}{4\pi\epsilon_0 r^2} = \frac{(-31\times 10^{-6})}{4\pi(8.8\times 10^{-12})(0.24)^2} = -4.86 \times 10^6~N/C$